Problem: $\begin{aligned} &H(x)=\left|\dfrac{x}{2}\right| \\\\ &h(x)=H'(x) \end{aligned}$ $\int_{-6}^{-2} h(x)\,dx=$
$h$ is the derivative of $H$, which means $H$ is an antiderivative of $h$. Since we know the antiderivative of $h$, we can use the fundamental theorem of calculus: For every function $h$ and its antiderivative $H$, $\int_a^b h(x)\,dx=H(b)-H(a)$. $\begin{aligned} &\phantom{=}\int_{-6}^{-2} h(x)\,dx \\\\ &=H({-2})-H({-6}) \\\\ &=\left|\dfrac{{-2}}{2}\right|-\left|\dfrac{{-6}}{2}\right| \\\\ &=1-3 \\\\ &=-2 \end{aligned}$ In conclusion, $\int_{-6}^{-2} h(x)\,dx=-2$